Question

Some of the steps in the derivation of the quadratic formula are shown.

Step 4: StartFraction negative 4 a c + b squared Over 4 a EndFraction = a ( x + StartFraction b Over 2 a EndFraction) squared

Step 5: (StartFraction 1 Over a EndFraction) StartFraction b squared minus 4 a c Over 4 a EndFraction = (StartFraction 1 Over a EndFraction) a (x + StartFraction b Over 2 a EndFraction) squared

Step 6: StartFraction b squared minus 4 a c Over 4 a squared EndFraction = ( x + StartFraction b Over 2 a EndFraction) squared

Step 7: StartFraction plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction = x + StartFraction b Over 2 a EndFraction
Which best explains why the expression plus or minus StartRoot b squared minus 4 a c EndRoot cannot be rewritten as b plus or minus StartRoot negative 4 a c EndRoot during the next step?

Negative values, like −4ac, do not have a square root.
The ± symbol prevents the square root from being evaluated.
The square root of terms separated by addition and subtraction cannot be calculated individually.
The entire term b2 − 4ac must be divided by 2a before its square root can be determined.

Answer 1

Answer:

(C)The square root of terms separated by addition and subtraction cannot be calculated individually.

Step-by-step explanation:

In Step 7:

$\dfrac{\pm\sqrt{b^2-4ac} }{2a} =x+\dfrac{b }{2a}$

The expression $\pm\sqrt{b^2-4ac}$ cannot be written as $b \pm\sqrt{4ac}$. This is as a result of the fact that square roots of terms separated by addition and subtraction cannot be calculated individually.

Take these examples:

$\sqrt{9-4} =\sqrt{5} \\\sqrt{9}-\sqrt{4}=3-2=1\\ Clearly \sqrt{5}\neq 1\\ Similarly\\\sqrt{9+4} =\sqrt{13} \\\sqrt{9}+\sqrt{4}=3+2=5\\ Clearly \sqrt{13}\neq 5$

Answer 2

Answer:

C

Step-by-step explanation:

Took the test on edge.